Question

For the following oxidation half reaction,
$\underset{\text{(Reducing ~agent)}}{2S_2{O}_3^{2-}}\to S_4{O}_6^{2-}+2e^{-}$
Calculate the equivalent weight of $S_2{O}_3^{2-}$.
(Given: $M$ is the molecular weight of $S_2{O}_3^{2-}$)

• A. $0.5M$
• B. $M$
• C. $1.5M$
• D. $2M$

Answer 1

The correct option is B $M$

The given equation is,
$2{\stackrel{+2}{S}}_2{O}_3^{2-}\to {\stackrel{+2.5}{S}}_4{O}_6^{2-}+2e^{-}$
formula used for the n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$
$n_f=|2-2.5|\times 2=1$
we know, $\text{Equivalent weight}=\frac{\text{Molecular weight}}{n_f}$
$\therefore \text{Equivalent weight of}{S}_2{O}_3^{2-}=\frac{M}{1}=M$